Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
Solution: It takes O(lgN) to find both the lower-bound and upper-bound.
1 class Solution {
2 public:
3 int lowerBound(int A[], int n, int target) {
4 int i = 0, j = n-1;
5 while(i <= j) {
6 int mid = i + (j-i)/2;
7 if(A[mid] < target)
8 i = mid+1;
9 else
10 j = mid-1;
11 }
12 if(i == -1 || i == n || A[i] != target) return -1;
13 else return i;
14 }
15
16 int upperBound(int A[], int n, int target) {
17 int i = 0, j = n-1;
18 while(i <= j) {
19 int mid = i + (j-i)/2;
20 if(A[mid] <= target) {
21 i = mid+1;
22 }
23 else {
24 j = mid-1;
25 }
26 }
27 if(j == -1 || j == n || A[j] != target) return -1;
28 else return j;
29 }
30
31 vector<int> searchRange(int A[], int n, int target) {
32 vector<int> res(2,-1);
33 res[0] = lowerBound(A, n, target);
34 res[1] = upperBound(A, n, target);
35 return res;
36 }
37 };